# Solution of first order nonlinear ODE

Can anyone please help me how to find the solution for: $$\frac{\mathrm dy}{\mathrm dx}=(x-5y)^{\frac{1}{3}}+ \frac{1}{5}.$$ I found the singular curve for it, which is $y=\dfrac{x}{5}$ (please correct me if I'm wrong), but I can not find the family of solution for it. I tried every method that I know. Please help me.

$$\frac{\mathrm dy}{\mathrm dx}=(x-5y)^{\frac{1}{3}}+ \frac{1}{5}.$$ $$5y'=5(x-5y)^{\frac{1}{3}}+1$$ $$-(1-5y')=5(x-5y)^{\frac{1}{3}}$$ You observe that the derivative of $x-5y$ is just $1-5y'$
$$\int \frac {d(x-5y)}{(x-5y)^{\frac{1}{3}}}=-5\int dx$$
It's easy to solve now $$\frac 32{(x-5y)^{\frac{2}{3}}}=-5x+K$$